The answer is negative. Indeed, for simplicity, let $a=-1$ and $b=1$.
In the case when $\delta=1/10$, $n=1$, and $X_1$ is uniformly distributed on $[-1,1]$ (so that $\mu=0$), let $f(\mathbf X,\delta):=1-1/10$. Then $ \Pr\left (\left | \mu -\frac{1}{n} \sum_{i=1}^n X_i \right | \leq f(\mathbf X, \delta) \right ) = \Pr(|X_1|\le1-1/10) = 1-1/10=1-\delta $, so that $(*)$ holds -- whereas $$ \Pr\left ( \mu -\frac{1}{n} \sum_{i=1}^n X_i \leq \frac{1}{2}f(\mathbf X, \delta) \right ) =\Pr\left(X_1\ge-\frac12\,(1-1/10)\right)=29/40 $$ $$\not\geq 1-1/10=1-\delta ,$$ so that $(**)$ fails to hold.
In all cases other than the one just considered, let e.g. $f(\mathbf X, \delta) := b-a$, so that $(*)$ hold.